How to do it...

1. Open src/Main.hs; we will add our prime number generator here.
2. We will start with 2 as the initial prime number and continue with only odd numbers; this will remove all the factors of 2.
3. We will assume that there are infinite prime numbers. We can write a list of primes as follows:

        primes :: [Integer]
        primes= 2 : filterMultiples allMultiples [3,5..]
        where
          allMultiples = mergeMultiples $ map multiples primes
          multiples i = map (i*) [i..]

Here, allMultiples are all multiples of all the primes; filterMultiples will weed out all those multiples from the list of odd numbers [3,5..]. All multiples are found out by lazily going over the primes that we are calculating and finding multiples of each.

4. We need to implement filterMultiples to weed out composites, merge to merge two list of multiples, and mergeMultiples to recursively merge all multiples of primes:

        filterMultiples :: Ord a => [a] -> [a] -> [a]
        filterMultiples (f:fs) (x:xs) | f < x      = filterMultiples fs
        (x:xs)
          | f > x      = x : filterMultiples (f:fs) xs
          | otherwise  = filterMultiples fs xs

        merge :: Ord a => [a] -> [a] -> [a]
        merge (x:xs) (y:ys) | x < y     = x : merge xs     (y:ys)
                    | x > y     = y : merge (x:xs) ys
                    | otherwise = x : merge xs     ys

       mergeMultiples :: Ord a => [[a]] -> [a]
       mergeMultiples ((x:xs):xss) = x : merge xs (mergeMultiples xss)

5. Finally, we can test our prime numbers by checking the 1000^th prime number (https://en.wikipedia.org/wiki/List_of_prime_numbers#The_first_1000_prime_numbers):

        main :: IO ()
        main = do
        let prime1k = take 1000 primes
        prime1kth = prime1k !! 999
        putStrLn $ "1000th prime number is " ++ (show prime1kth)

6. If you build the project and run it, then you should see the 1000^th prime number: